The image presents a set of problems related to systems of equations. These include solving systems by graphing, solving systems algebraically, verifying if a point is a solution to a system, finding the number of solutions for a system, and solving a word problem involving a system of equations.

1. Problem Description

The image presents a set of problems related to systems of equations. These include solving systems by graphing, solving systems algebraically, verifying if a point is a solution to a system, finding the number of solutions for a system, and solving a word problem involving a system of equations.

2. Solution Steps

Problem 2: Solve the system

$4x - 5y = -3$

$9x + y = 19$

We can solve this system using substitution or elimination. Let's use substitution. From the second equation, we have

$y = 19 - 9x$

Substitute this expression for $y$ into the first equation:

$4x - 5(19 - 9x) = -3$

$4x - 95 + 45x = -3$

$49x = 92$

$x = \frac{92}{49}$

Now, substitute this value of $x$ back into the equation for $y$:

$y = 19 - 9(\frac{92}{49})$

$y = 19 - \frac{828}{49}$

$y = \frac{19 \cdot 49 - 828}{49}$

$y = \frac{931 - 828}{49}$

$y = \frac{103}{49}$

Problem 3: Is (5, -2) a solution of this system?

$5x + y = 23$

$56 = 12x - 2y$

Substitute $x=5$ and $y=-2$ into the first equation:

$5(5) + (-2) = 25 - 2 = 23$. This is true.

Substitute $x=5$ and $y=-2$ into the second equation:

$12(5) - 2(-2) = 60 + 4 = 64$. This is not equal to

5

6.

Therefore, (5, -2) is not a solution of the system.

Problem 4: Find the number of solutions.

$8x - 2y + 10 = 0$

$y = 4x + 5$

Substitute the second equation into the first equation:

$8x - 2(4x + 5) + 10 = 0$

$8x - 8x - 10 + 10 = 0$

$0 = 0$

Since we get an identity ($0 = 0$), this means the two equations are dependent, and there are infinitely many solutions.

Problem 5:

Let $s$ be the number of student tickets and $a$ be the number of adult tickets.

$s + a = 455$

$15s + 18a = 6924$

From the first equation, $s = 455 - a$. Substitute this into the second equation:

$15(455 - a) + 18a = 6924$

$6825 - 15a + 18a = 6924$

$3a = 99$

$a = 33$

Now find the number of students:

$s = 455 - 33 = 422$

3. Final Answer

Problem 2: $x = \frac{92}{49}$, $y = \frac{103}{49}$

Problem 3: No, (5, -2) is not a solution.

Problem 4: Infinitely many solutions.

Problem 5: 422 students and 33 teachers.